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Let $f$ be a function defined by : $$\forall x\in(0,+\infty)\quad f(x)=x+\dfrac{1}{2}\dfrac{\ln(x)}{x} $$

calculate $f^{-1}(1)$

let $\alpha \in(0,+\infty) $ such that $f^{-1}(1)=\alpha$ then $f(\alpha)=1 \implies 2\alpha(\alpha-1)+\ln( \alpha)=0$ i'm stuck here i can't solve this equation i know that the solution is $1$ could we solve determinate inverse function of f i.e. $f^{-1}$

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    i think you can not calculate $$f^{-1}(x)$$ in an explicit form – 2017-01-09
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    that's what i thought – 2017-01-09
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    `i know that the solution is 1` That's correct, and $f(1)=1$ can be easily verified by direct substitution. This implies $f^{-1}(1)=1$ which answers the question as posted. – 2017-01-09

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For this special case, you can solve by elimination: $$ \begin{align} & 2\alpha(\alpha-1)+\log(\alpha)=0 \implies \log(\alpha)=2\alpha(1-\alpha) \\[2mm] & \text{Let }\space g(\alpha)=\log(\alpha) \quad\&\quad h(\alpha)=2\alpha(1-\alpha) \\[2mm] & \qquad \begin{cases} \alpha \,\in\, ]0,1[ &\Rightarrow\quad g(\alpha)=\log(\alpha) \lt0 \,,\space\space h(\alpha)=2\alpha(1-\alpha) \gt0 \\ \alpha \,\in\, ]1,+\infty[ &\Rightarrow\quad g(\alpha)=\log(\alpha) \gt0 \,,\space\space h(\alpha)=2\alpha(1-\alpha) \lt0 \end{cases} \\[2mm] & \qquad \implies \color{red}{g(\alpha) \ne h(\alpha) \space\colon\space \alpha \,\in\, ]0,+\infty[\,-\,\{1\}} \\[2mm] & \text{Remaining only }\space \color{red}{\alpha=1} \Rightarrow g(1)=\log(1)=0 \space\&\space h(1)=2\times1(1-1)=0 \\[2mm] & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \end{align} $$ enter image description here